Verifying that $ prod_{j=1}^{infty} frac{1}{1-q^j} = prod_{j=1}^{infty} frac{1}{(1-q^{2j-1})(1-q^{2j})}$












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On page 165 of Chapter 13, how was the equality made from line 1 to line 2?



https://archive.org/details/NumberTheory_862/page/n173



Namely, how $$ prod_{j=1}^{infty} frac{1}{1-q^j} = prod_{j=1}^{infty} frac{1}{(1-q^{2j-1})(1-q^{2j})}$$






infinite-product integer-partitions







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    0












    $begingroup$


    On page 165 of Chapter 13, how was the equality made from line 1 to line 2?



    https://archive.org/details/NumberTheory_862/page/n173



    Namely, how $$ prod_{j=1}^{infty} frac{1}{1-q^j} = prod_{j=1}^{infty} frac{1}{(1-q^{2j-1})(1-q^{2j})}$$






    infinite-product integer-partitions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      On page 165 of Chapter 13, how was the equality made from line 1 to line 2?



      https://archive.org/details/NumberTheory_862/page/n173



      Namely, how $$ prod_{j=1}^{infty} frac{1}{1-q^j} = prod_{j=1}^{infty} frac{1}{(1-q^{2j-1})(1-q^{2j})}$$






      infinite-product integer-partitions







      share|cite|improve this question











      $endgroup$




      On page 165 of Chapter 13, how was the equality made from line 1 to line 2?



      https://archive.org/details/NumberTheory_862/page/n173



      Namely, how $$ prod_{j=1}^{infty} frac{1}{1-q^j} = prod_{j=1}^{infty} frac{1}{(1-q^{2j-1})(1-q^{2j})}$$






      infinite-product integer-partitions




      infinite-product integer-partitions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 17 '18 at 0:42









      Blue

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      asked Dec 17 '18 at 0:25









      zodrosszodross

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          1 Answer
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          $begingroup$

          Simply pair-off factors and use the fact that multiplication is commutative:
          begin{align*}
          prod_{j=1}^{infty}frac{1}{(1-q^{j})} & = frac{1}{1-q}frac{1}{1-q^2}frac{1}{1-q^3}frac{1}{1-q^4}cdots\
          & = left(frac{1}{1-q}frac{1}{1-q^2}right)left(frac{1}{1-q^3}frac{1}{1-q^4}right)cdots\
          & = left(frac{1}{1-q^{2}}frac{1}{1-q}right)left(frac{1}{1-q^4}frac{1}{1-q^{3}}right)cdots\
          & = prod_{j=1}^{infty}frac{1}{(1-q^{2j})(1-q^{2j-1})}.
          end{align*}






          share|cite|improve this answer









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          • $begingroup$
            Oh, I see. Thank you.
            $endgroup$
            – zodross
            Dec 17 '18 at 0:32











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          $begingroup$

          Simply pair-off factors and use the fact that multiplication is commutative:
          begin{align*}
          prod_{j=1}^{infty}frac{1}{(1-q^{j})} & = frac{1}{1-q}frac{1}{1-q^2}frac{1}{1-q^3}frac{1}{1-q^4}cdots\
          & = left(frac{1}{1-q}frac{1}{1-q^2}right)left(frac{1}{1-q^3}frac{1}{1-q^4}right)cdots\
          & = left(frac{1}{1-q^{2}}frac{1}{1-q}right)left(frac{1}{1-q^4}frac{1}{1-q^{3}}right)cdots\
          & = prod_{j=1}^{infty}frac{1}{(1-q^{2j})(1-q^{2j-1})}.
          end{align*}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh, I see. Thank you.
            $endgroup$
            – zodross
            Dec 17 '18 at 0:32
















          2












          $begingroup$

          Simply pair-off factors and use the fact that multiplication is commutative:
          begin{align*}
          prod_{j=1}^{infty}frac{1}{(1-q^{j})} & = frac{1}{1-q}frac{1}{1-q^2}frac{1}{1-q^3}frac{1}{1-q^4}cdots\
          & = left(frac{1}{1-q}frac{1}{1-q^2}right)left(frac{1}{1-q^3}frac{1}{1-q^4}right)cdots\
          & = left(frac{1}{1-q^{2}}frac{1}{1-q}right)left(frac{1}{1-q^4}frac{1}{1-q^{3}}right)cdots\
          & = prod_{j=1}^{infty}frac{1}{(1-q^{2j})(1-q^{2j-1})}.
          end{align*}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh, I see. Thank you.
            $endgroup$
            – zodross
            Dec 17 '18 at 0:32














          2












          2








          2





          $begingroup$

          Simply pair-off factors and use the fact that multiplication is commutative:
          begin{align*}
          prod_{j=1}^{infty}frac{1}{(1-q^{j})} & = frac{1}{1-q}frac{1}{1-q^2}frac{1}{1-q^3}frac{1}{1-q^4}cdots\
          & = left(frac{1}{1-q}frac{1}{1-q^2}right)left(frac{1}{1-q^3}frac{1}{1-q^4}right)cdots\
          & = left(frac{1}{1-q^{2}}frac{1}{1-q}right)left(frac{1}{1-q^4}frac{1}{1-q^{3}}right)cdots\
          & = prod_{j=1}^{infty}frac{1}{(1-q^{2j})(1-q^{2j-1})}.
          end{align*}






          share|cite|improve this answer









          $endgroup$



          Simply pair-off factors and use the fact that multiplication is commutative:
          begin{align*}
          prod_{j=1}^{infty}frac{1}{(1-q^{j})} & = frac{1}{1-q}frac{1}{1-q^2}frac{1}{1-q^3}frac{1}{1-q^4}cdots\
          & = left(frac{1}{1-q}frac{1}{1-q^2}right)left(frac{1}{1-q^3}frac{1}{1-q^4}right)cdots\
          & = left(frac{1}{1-q^{2}}frac{1}{1-q}right)left(frac{1}{1-q^4}frac{1}{1-q^{3}}right)cdots\
          & = prod_{j=1}^{infty}frac{1}{(1-q^{2j})(1-q^{2j-1})}.
          end{align*}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 0:30









          Will RWill R

          6,59731429




          6,59731429












          • $begingroup$
            Oh, I see. Thank you.
            $endgroup$
            – zodross
            Dec 17 '18 at 0:32


















          • $begingroup$
            Oh, I see. Thank you.
            $endgroup$
            – zodross
            Dec 17 '18 at 0:32
















          $begingroup$
          Oh, I see. Thank you.
          $endgroup$
          – zodross
          Dec 17 '18 at 0:32




          $begingroup$
          Oh, I see. Thank you.
          $endgroup$
          – zodross
          Dec 17 '18 at 0:32


















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